Question

The half-life of a radioactive isotope $\text{X}$ is $20\text{years}.$ It decays into another element $\text{Y}$, which is stable. The two elements $\text{X}$ and $\text{Y}$ were found to be in the ratio $1:7$, in a sample of a given rock. The age of the rock is estimated to be,

• A. $100\text{years}$
• B. $40\text{years}$
• C. $60\text{years}$
• D. $80\text{years}$

Answer 1

The correct option is C $60\text{years}$

Given, $\frac{N_X}{N_Y}=\frac{1}{7}$

$\text{X}\to \text{Y (stable)}$

Initially, $N_0=N_X+N_Y$

Now, after $t$ years, $N_X=N$

$\therefore \frac{N}{N_0}=\frac{N_X}{N_X+N_Y}=\frac{1}{8}={\left(\frac{1}{2}\right)}^3$

$\Rightarrow N=\frac{N_0}{8}$

Using law of radioactivity,

$N=N_0{e}^{-\lambda t}$

$\frac{N}{N_0}=e^{-\lambda t}$

$8=e^{\lambda t}$

$\ln 8=\lambda t$

$\ln 8=\frac{\ln 2}{t_{1/2}}t$

$3\ln 2=\frac{\ln 2}{t_{1/2}}t$

$\Rightarrow t=3t_{1/2}=60\text{years}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.